I remember back when I was still an engineering student circa 2009 i used to rely on khan academy math tutorials, just wished there were more resources like this back in the day
When I first looked at it I did the exact thing which was 9^x = 10^x which ovb doesn't have any solution ( I accidentally added them to 10^x my bad sorry) 9^x = 4^x+6^x
@@AshishSinghGAMING 10^x ?? Its addition, not multiplication. Also, 9^x = 10^x has a solution. I leave it up to you to figure that out. It is easy though.
@@AshishSinghGAMING You can't add numbers raised to powers that way. 5^2 = 3^2 + 4^2, or 25=9+16. But 5^2 is not equal to (3+4)^2. (3+4)^2 is like (a+b)^2, which is a^2 + 2ab + b^2. The answer is larger than a^2 + b^2 by the value 2ab. (3+4)^2 is 49, which is 24 more than 9+16. 24=2*3*4, right there is the 2ab!.
Nice Challenge. Please state in the beginning (!), that you want real solutions, only. The negative solution of the quadratic equation leads to a complex solution of the problem. This complex solution does solve the problem, also !
I have a feeling, he wanted to show off all logarithmic laws in this video. Because there were faster ways and the final step literally does not do anything, but the steps were necessary to use the laws. Still fun to watch and easy to follow. Keep it up
taking the log in base (3/2) to change as a log in another base.... interesting but you spend less time in taking the log in normal base. Nethertheless, this is a good teaching solution for students and very careful (excuse my english)
for a start you can show it has a solution (whatever it may be) pretty easy: 9^x-6^x=4^x | +6^x 9^x=4^x+6^x in this case x must be less than 1 as 4^1+6^1 would be 10^1
Not really 'shown'. If you move everything to one side equal to zero you can show that at 0 the function is negative and at 1 it it positive. Then, given that the function is continuous it must have a solution between 0 and 1. You missed showing any lower bound or stating that the function is continuous.
My solution: 9^x-6^x=4^x =>9^x-6^x+4^(x-1)=4^x+4^(x-1) =>(3^x-2^(x-1))²=4^x(1+1/4) =>(3^x-2^(x-1)²/(4^x)=5/4 =>(1,5^x-1/2)²=5/4 =>1,5^x-1/2=√(5/4) (negative root makes no sense to use here so only do positive) =>x=log(1,5) (√(5/4)+1/2)
Put in calculator 9x1.187 multiply by 6x1.187 and the answer won't even be 4x1.187. I already knew the answer was less than one before using caluclor. Guys please correct me if I'm wrong😅
Im an Engineer, working for 6 years now but still find this interesting 🎉
I remember back when I was still an engineering student circa 2009 i used to rely on khan academy math tutorials, just wished there were more resources like this back in the day
How is this supposed to be a high school entrance exam if that is all content that is being tsught in high school? Great video btw.
Exactly what I thought.
Private high school
@@wanderingfido nope
My school used to give questions like this. The school didn't expect students to answer it, but specially rewarded those that were able to.
Just a note: Rewriting the problem as 9^x = 6^x + 4^x cannot have any integer solutions > 2 according to Fermat's last theorem.
When I first looked at it I did the exact thing which was 9^x = 10^x which ovb doesn't have any solution ( I accidentally added them to 10^x my bad sorry) 9^x = 4^x+6^x
But can you tell me more about Fermat's last theorem
Just odd even numbers are enough to support your statement.
@@AshishSinghGAMING 10^x ?? Its addition, not multiplication. Also, 9^x = 10^x has a solution. I leave it up to you to figure that out. It is easy though.
@@AshishSinghGAMING You can't add numbers raised to powers that way. 5^2 = 3^2 + 4^2, or 25=9+16. But 5^2 is not equal to (3+4)^2.
(3+4)^2 is like (a+b)^2, which is a^2 + 2ab + b^2. The answer is larger than a^2 + b^2 by the value 2ab. (3+4)^2 is 49, which is 24 more than 9+16. 24=2*3*4, right there is the 2ab!.
No mention that (1+sqrt 5)/2 is the golden ratio?!?! φ never gets the respect it deserves.
Simpler: divide by 9^x. Set z= (2/3)^x. Then z^2 -z+1=0. So (2/3)^x = the z solution.
I graphed this on Desmos and got a more accurate answer of 1.18681.
I'm really glad I'm retired and don't have to take exams to get into high school.
Nice Challenge. Please state in the beginning (!), that you want real solutions, only. The negative solution of the quadratic equation leads to a complex solution of the problem. This complex solution does solve the problem, also !
Interesting question, thank you very much 🙏
If t^2 - t - 1 = 0, then t = φ or -1/φ.
I wish more UA-camr mathematicians would use φ for (1+sqrt(5))/2.
Problem is, unless you get farther into math, or just like to self teach, phi is not a very well known constant
very good problem anf your expression simplifes the the question. i appreciated it as an old diferential and integral lessons teacher.
Minor error: log(b,c)=log(m,c)/log(m,b), not log(b,c)=log(m,a)/log(m,b).
Otherwise great video!
At first I was confused as well but Ig it was just a typo
8:43 Using the natural logarithm gives the answer more directly
I have a feeling, he wanted to show off all logarithmic laws in this video. Because there were faster ways and the final step literally does not do anything, but the steps were necessary to use the laws. Still fun to watch and easy to follow. Keep it up
Pretty sure this is a university entrance exam, not high school.
This is supposed to be part of a High School entrance exam?
taking the log in base (3/2) to change as a log in another base.... interesting but you spend less time in taking the log in normal base. Nethertheless, this is a good teaching solution for students and very careful (excuse my english)
9x - 6x = 4x
3x = 4x
x = log 4 / log 3
x = 1.26186
3^1.26186 = 4
This gave me anxiety and depression
Respected Sir, Good evening....❤
Minus infinity. Done.
I mean I guess that kinda works 😭
Can log be taken at initial stage
I did it in my head.
Same.
I wanna learn that shortcut, getting back into maths
Congrats??
for a start you can show it has a solution (whatever it may be) pretty easy:
9^x-6^x=4^x | +6^x
9^x=4^x+6^x
in this case x must be less than 1 as 4^1+6^1 would be 10^1
Not really 'shown'. If you move everything to one side equal to zero you can show that at 0 the function is negative and at 1 it it positive. Then, given that the function is continuous it must have a solution between 0 and 1. You missed showing any lower bound or stating that the function is continuous.
My solution:
9^x-6^x=4^x
=>9^x-6^x+4^(x-1)=4^x+4^(x-1)
=>(3^x-2^(x-1))²=4^x(1+1/4)
=>(3^x-2^(x-1)²/(4^x)=5/4
=>(1,5^x-1/2)²=5/4
=>1,5^x-1/2=√(5/4) (negative root makes no sense to use here so only do positive)
=>x=log(1,5) (√(5/4)+1/2)
(9)^2=81(4)^2,=16 {81 ➖ 16}= 75 3^25 3^5^5 3^1^1 3^1 (x ➖ 3x+1). (4)^2= 16 4^4 2^2^2^2 1^1^1^2 1^2 (x ➖ 2x+1).
9^x-6^x=4^x
(3/2)^(2x)-(3/2)^x=1
u=(3/2)^x → u²-u-1=0
u=(1±√5)/2
u>0 → u=(1+√5)/2
∴(3/2)^x=(1+√5)/2
x=log{(1+√5)/2}/log(3/2)
Please don't t write 1 the way you write. It looks like 4. Simply write 1 as I . Thank you
/:4^x , (9/4)^x-(3/2)^x=1 , let u=(3/2)^x , u^2-u-1=0 , u= (1+V5)/2 , / (1-V5)/2 < 0 not a solu / , (3/2)^x= (1+V5)/2 ,
x=ln((1+V5)/2)/ln(3/2) , x=~ 1.18681... ,
test , 9^(ln((1+V5)/2)/ln(3/2))-6^(ln((1+V5)/2)/ln(3/2))=~ 5.18243 , 4^(ln((1+V5)/2)/ln(3/2))=~ 5.18243 , same , OK ,
You write too small. Very hard to see
You got the wrong answer I checked by calculator the real answer is less than 1. I'm in 8th grade and I know this
Put in calculator 9x1.187 multiply by 6x1.187 and the answer won't even be 4x1.187. I already knew the answer was less than one before using caluclor. Guys please correct me if I'm wrong😅
Solved this in less than 15 seconds without even writing anything down.
!!!!!!!
Input interpretation
9^(log(1.5, 0.3^_ (repeating decimal) (1 + sqrt(5))) + 1) - 6^(log(1.5, 0.3^_ (repeating decimal) (1 + sqrt(5))) + 1) = 4^(log(1.5, 0.3^_ (repeating decimal) (1 + sqrt(5))) + 1)
Result
True
Logarithmic form
log(4, 9^(log(1.5, 0.3^_ (repeating decimal) (1 + sqrt(5))) + 1) - 6^(log(1.5, 0.3^_ (repeating decimal) (1 + sqrt(5))) + 1)) = (log(1.5, 0.3^_ (repeating decimal) (1 + sqrt(5))) + 1) log(4, 4)
Pretentious mathematics.
You lost me!